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ietr said:Obviously there is some error in the above three equations because there are four unknowns: T, a1, a2 and a3.
I'm having trouble understanding this:
* Will the tension remain T along all segments of the string, unlike what I have assumed above?
* Since the string length is constant, the accelerations a1, a2 and a3 must be equal but if we set them all equal to 'a', then we have only two unknowns T and 'a' and three equations, and further on solving them we do not get the correct answer.
Orodruin said:If instead of being inclined by 30 degrees, the thread would be hanging vertically with a 100 g weight in the middle, would the tension be the same above and below the weight?
I guess that yes the tension would be T in both the sides of the weight
Yes, the accelerations have to be of equal magnitude - but be very mindful of their direction, if one of the ends accelerates up, the other end accelerates down.
It's not clear from your diagram how the 100g block is attached. Looks to like there are (logically, at least) two separate strings - one runs from the 500g block, over the top pulley, to the 100g block; the second runs from the 100g block, over the lower pulley, to the 50g block. There is no reason why the tensions should be the same in the two sections of string.ietr said:if a1 = a2 = a3, and we assume that their directions are:
a1 : of 500 g downwards
a2: of 100 g upwards along the incline
a3: of 50 g upwards
then the equations become (assuming that the tension is T along the length of the string)
(1/2)g - T = (1/2)a
T - (1/10)g sin(30) = (1/10)a
T - (1/20)g = (1/20)a
Solving these equations for 'a', we get a = (3/4)g from the first two equations and a = (9/11)g from the first and third equations. How can there be two values of 'a'?
haruspex said:It's not clear from your diagram how the 100g block is attached. Looks to like there are (logically, at least) two separate strings - one runs from the 500g block, over the top pulley, to the 100g block; the second runs from the 100g block, over the lower pulley, to the 50g block. There is no reason why the tensions should be the same in the two sections of string.
No, it does matter. Within the range of possibilities in the diagram, the 100g block would tip over, even if there is no friction.Orodruin said:It does not matter much how it is attached. What matters is that it is attached (and parallel to the inlined plane).
Solving a two pulley problem involves using the principles of physics, specifically torque and rotational equilibrium. The first step is to draw a free body diagram and label all the forces acting on the pulleys and the objects they are connected to. Then, use the equations τ = rF and Στ = 0 to find the unknown forces and tensions.
A fixed pulley is attached to a stationary object and only changes the direction of the applied force. A movable pulley is attached to the object being lifted and changes both the direction and magnitude of the applied force. In a two pulley problem, both types of pulleys are often used together.
The correct way to set up a two pulley problem is to draw a diagram with the pulleys and objects, label all the forces acting on them, and write down the known and unknown values. It is important to choose a coordinate system and stick to it throughout the problem.
To check if your answer is correct, you can use the equations ΣF = ma and Στ = Iα to ensure that the forces and torques are balanced. You can also compare your answer to the expected outcome, such as the weight of the object or the tension in the rope, to see if they are reasonable.
Yes, there are multiple methods for solving two pulley problems, such as using vector analysis, energy conservation, or the concept of work. It is important to choose the method that you are most comfortable with and that will yield the most accurate and efficient solution.